a,\(|x\left(x-4\right)|=x\)
b,\(|x+2|-6x=1\)
c, \(|x^2-2x|=x\)
Những câu hỏi liên quan
16 tháng 4 2023 lúc 22:38
\(A\left(x\right)=2x^3+4x^2-x-1\)
+
\(B\left(x\right)=2x^3-2x^2-x-3\)
=\(P\left(x\right)=4x^3-6x^2-4\)
(a): đúng
(b): sai
\(A\left(x\right)+B\left(x\right)=2x^3+4x^2-x-1+2x^3-2x^2-x-3\\ =\left(2x^3+2x^3\right)+\left(4x^2-2x^2\right)+\left(-x-x\right)+\left(-1-3\right)\\ =4x^3+2x^2-2x-4\ne P\left(x\right)\)
=> Chọn B. Sai
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tìm x biết
a)\(\left(x-2\right)^2-x\left(x+1\right)\left(x-1\right)+6x\left(x-3\right)=0\)
b)\(\left(x-2\right)\left(x^2-2x+4\right)\left(x+2\right)\left(x^2+2x+4\right)-x^6+2x=1\)
c)\(\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\)
b) Ta có: \(\left(x-2\right)\left(x^2-2x+4\right)\left(x+2\right)\left(x^2+2x+4\right)-x^6+2x=1\)
\(\Leftrightarrow\left(x^3-8\right)\left(x^3+8\right)-x^6+2x-1=0\)
\(\Leftrightarrow x^6-64-x^6+2x-1=0\)
\(\Leftrightarrow2x-65=0\)
\(\Leftrightarrow2x=65\)
hay \(x=\frac{65}{2}\)
Vậy: \(x=\frac{65}{2}\)
c) Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\)
\(\Leftrightarrow x^3-27-x\left(x+2\right)\left(x-2\right)-1=0\)
\(\Leftrightarrow x^3-27-x\left(x^2-4\right)-1=0\)
\(\Leftrightarrow x^3-27-x^3+4x-1=0\)
\(\Leftrightarrow4x-28=0\)
\(\Leftrightarrow4x=28\)
hay x=7
Vậy: x=7
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Giải các phương trình sau:
a) x^2+dfrac{2x}{x-1}8
b) dfrac{x^2+2x+1}{x^2+2x+2}+dfrac{x^2+2x+2}{x^2+2x+3}dfrac{7}{6}
c) dfrac{x+4}{x-1}+dfrac{x-4}{x+1}dfrac{x+8}{x-2}+dfrac{x-8}{x+2}+6
d) left(x^2+6x+8right)left(x^2+8x+15right)24
e) left(x^2+x-2right)left(x^2+9x+18right)28
f) 3left(-x^2+2x+3right)^4-26x^2left(-x^2+2x+3right)^2-9x^40
g) x^4+6x^3+11x^2+6x+10
h) left(x-3right)left(x-5right)left(x-6right)left(x-10right)-24x^20
i) left(x+2right)^4+left(x+8right)^4272
Đọc tiếp
Giải các phương trình sau:
a) \(x^2+\dfrac{2x}{x-1}=8\)
b) \(\dfrac{x^2+2x+1}{x^2+2x+2}+\dfrac{x^2+2x+2}{x^2+2x+3}=\dfrac{7}{6}\)
c) \(\dfrac{x+4}{x-1}+\dfrac{x-4}{x+1}=\dfrac{x+8}{x-2}+\dfrac{x-8}{x+2}+6\)
d) \(\left(x^2+6x+8\right)\left(x^2+8x+15\right)=24\)
e) \(\left(x^2+x-2\right)\left(x^2+9x+18\right)=28\)
f) \(3\left(-x^2+2x+3\right)^4-26x^2\left(-x^2+2x+3\right)^2-9x^4=0\)
g) \(x^4+6x^3+11x^2+6x+1=0\)
h) \(\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)-24x^2=0\)
i) \(\left(x+2\right)^4+\left(x+8\right)^4=272\)
giải hết đống này chắc @@ quá,để tối đi,giờ t đi làm mấy bài ngắn ngắn
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a) x2+\(\dfrac{2x}{x-1}\)=8(ĐKXĐ : x ≠ 1
⇔ x2(x-1)+2x =8⇔ x3 - x2 +2x - 8=0
⇔x3 - 23 -x2+2x =0⇔ (x-2)(x2 +x+1) -x(x-2)
⇔(x-2)(x2 +1)⇒x =2
x2 +1 =0⇒x2 -1⇒x ∈∅(loại)
vậy x =2
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Giải phương trình :
a) \(x^2+\dfrac{2x}{x-1}=8\)
ĐKXĐ : \(x-1\ne0\Rightarrow x\ne1\)
Ta có : \(x^2+\dfrac{2x}{x-1}=8\)
\(\Leftrightarrow\) \(\dfrac{x^2\left(x-1\right)}{x-1}+\dfrac{2x}{x-1}=\dfrac{8\left(x-1\right)}{x-1}\)
\(\Rightarrow x^2\left(x-1\right)+2x=8\left(x-1\right)\)
\(\Leftrightarrow x^3-x^2+2x=8x-8\)
\(\Leftrightarrow x^3-x^2+2x-8x=-8\)
\(\Leftrightarrow x^3-x^2-6x+8=0\)
\(\Leftrightarrow\left(x^3-x^2\right)-\left(6x-8\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)-2\left(3x-4\right)=0\)
\(\Leftrightarrow\left(x^2-2\right)\left(x-1\right)\left(3x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2=0\\x-1=0\\3x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=1\\x=\dfrac{4}{3}\end{matrix}\right.\)
Đối chiếu với ĐKXĐ ta được \(x\in\left\{\sqrt{2};\dfrac{4}{3}\right\}\) thỏa mãn.
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Tìm GTNN của các biểu thức:
a) Aleft(x+8right)^4+left(x+6right)^4
b) Bleft(0,5x^2+xright)^2-3left|0,5x^2+xright|
c) Cleft(x-1right)left(x-3right)left(x^2-4x+5right)
d) Dx^4-2x^3+3x^2-2x+1
e) E x^4-6x^3+10x^2-6x+9
g) Gleft(x^2+x-6right)left(x^2+x+2right)
Đọc tiếp
Tìm GTNN của các biểu thức:
a) \(A=\left(x+8\right)^4+\left(x+6\right)^4\)
b) \(B=\left(0,5x^2+x\right)^2-3\left|0,5x^2+x\right|\)
c) \(C=\left(x-1\right)\left(x-3\right)\left(x^2-4x+5\right)\)
d) \(D=x^4-2x^3+3x^2-2x+1\)
e) E = \(x^4-6x^3+10x^2-6x+9\)
g) \(G=\left(x^2+x-6\right)\left(x^2+x+2\right)\)
a) x-dfrac{dfrac{x}{2}-dfrac{3+x}{4}}{2}dfrac{2x-dfrac{10-7x}{3}}{3}-left(x-1right)
b) x^2-6x-2+dfrac{14}{x^2-6x+7}0
c) dfrac{8x^2}{3left(1-4x^2right)}dfrac{2x}{6x-3}-dfrac{1+8x}{4+8x}
d) dfrac{13}{left(2x+7right)left(x-3right)}+dfrac{1}{left(2x+7right)}dfrac{6}{x^2-9}
e) left(1-dfrac{2x-1}{x+1}right)^3+6left(1-dfrac{2x-1}{x+1}right)^2dfrac{12left(2x-1right)}{x+1}-20
Đọc tiếp
a) \(x-\dfrac{\dfrac{x}{2}-\dfrac{3+x}{4}}{2}=\dfrac{2x-\dfrac{10-7x}{3}}{3}-\left(x-1\right)\)
b) \(x^2-6x-2+\dfrac{14}{x^2-6x+7}=0\)
c) \(\dfrac{8x^2}{3\left(1-4x^2\right)}=\dfrac{2x}{6x-3}-\dfrac{1+8x}{4+8x}\)
d) \(\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}=\dfrac{6}{x^2-9}\)
e) \(\left(1-\dfrac{2x-1}{x+1}\right)^3+6\left(1-\dfrac{2x-1}{x+1}\right)^2=\dfrac{12\left(2x-1\right)}{x+1}-20\)
b: Đặt \(x^2-6x-2=a\)
Theo đề, ta có: \(a+\dfrac{14}{a+9}=0\)
=>(a+2)(a+7)=0
\(\Leftrightarrow\left(x^2-6x\right)\left(x^2-6x+5\right)=0\)
=>x(x-6)(x-1)(x-5)=0
hay \(x\in\left\{0;1;6;5\right\}\)
c: \(\Leftrightarrow\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)
\(\Leftrightarrow-32x^2=8x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)\)
\(\Leftrightarrow-32x^2=16x^2+8x-3\left(16x^2-8x+2x-1\right)\)
\(\Leftrightarrow-48x^2=8x-48x^2+18x+3\)
=>26x=-3
hay x=-3/26
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25 tháng 1 2021 lúc 22:07
Giair phương trình sau:a,2x^3+5x^2-3x0 b,2x^3+6x^2x^2+3xc,x^2+left(x+2right)left(11x-7right)4 d,left(x-1right)left(x^2+5x-2right)-left(x^3-1right)0e, x^3+1xleft(x+1right) f,x^3+x^2+x+10g,x^3-3x^2+3x-10 h,x^3-7x+60i,x^6-x^20 j,x^3-1213xk,-x^5+4x^4-12x^3 l, x^34x
Đọc tiếp
Giair phương trình sau:
a,\(2x^3+5x^2-3x=0\) b,\(2x^3+6x^2=x^2+3x\)
c,\(x^2+\left(x+2\right)\left(11x-7\right)=4\) d,\(\left(x-1\right)\left(x^2+5x-2\right)-\left(x^3-1\right)=0\)
e, \(x^3+1=x\left(x+1\right)\) f,\(x^3+x^2+x+1=0\)
g,\(x^3-3x^2+3x-1=0\) h,\(x^3-7x+6=0\)
i,\(x^6-x^2=0\) j,\(x^3-12=13x\)
k,\(-x^5+4x^4=-12x^3\) l, \(x^3=4x\)
a) Ta có: \(2x^3+5x^2-3x=0\)
\(\Leftrightarrow x\left(2x^2+5x-3\right)=0\)
\(\Leftrightarrow x\left(2x^2+6x-x-3\right)=0\)
\(\Leftrightarrow x\left[2x\left(x+3\right)-\left(x+3\right)\right]=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)
b) Ta có: \(2x^3+6x^2=x^2+3x\)
\(\Leftrightarrow2x^2\left(x+3\right)=x\left(x+3\right)\)
\(\Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)
c) Ta có: \(x^2+\left(x+2\right)\left(11x-7\right)=4\)
\(\Leftrightarrow x^2+11x^2-7x+22x-14-4=0\)
\(\Leftrightarrow12x^2+15x-18=0\)
\(\Leftrightarrow12x^2+24x-9x-18=0\)
\(\Leftrightarrow12x\left(x+2\right)-9\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(12x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\12x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\12x=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{-2;\dfrac{3}{4}\right\}\)
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Trong đó có nhiều phương trình kiến thức cơ bản mà nhỉ? Ít nâng cao, bạn lọc ra câu nào k làm đc thôi chứ!
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a,\(\dfrac{x+1}{x-3}+\dfrac{-2x^2+2x}{x^2-9}+\dfrac{x-1}{x+3}\)
b,\(\dfrac{1-2x}{6x^3y}+\dfrac{3+2y}{6x^3y}+\dfrac{2x-4}{6x^3y}\)
c,\(\dfrac{5}{2x^2y}+\dfrac{3}{5xy^2}+\dfrac{x}{3y^3}\)
d,\(\dfrac{5}{4\left(x+2\right)}+\dfrac{8-x}{4x^2+8x}\)
c,\(\dfrac{x^2+2}{x^3+1}+\dfrac{2}{x^2+x+1}+\dfrac{1}{1-x}\)
\(a,=\dfrac{x^2+4x+3-2x^2+2x+x^2-4x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ b,=\dfrac{1-2x+3+2y+2x-4}{6x^3y}=\dfrac{2y}{6x^3y}=\dfrac{1}{x^2}\\ c,=\dfrac{75y^2+18xy+10x^2}{30x^2y^3}\\ d,=\dfrac{5x+8-x}{4x\left(x+2\right)}=\dfrac{4\left(x+2\right)}{4x\left(x+2\right)}=\dfrac{1}{x}\\ c,=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)
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Bài 2 . Thực hiện phép tính
a)\(6x^3\)\(\left(\dfrac{1}{3}x^2-\dfrac{5}{2}-\dfrac{1}{6}\right)\)\(-2x^5\)\(-x^3\)
b)\(\left(x-3\right)\left(x^2+3x-2\right)\)
c)\(\left(4x^3-4x^2-5x+4\right):\left(2x+1\right)\)
a: =2x^5-15x^3-x^2-2x^5-x^3=-16x^3-x^2
b: =x^3+3x^2-2x-3x^2-9x+6
=x^3-11x+6
c: \(=\dfrac{4x^3+2x^2-6x^2-3x-2x-1+5}{2x+1}\)
\(=2x^2-3x-1+\dfrac{5}{2x+1}\)
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a) \(6x^3\left(\dfrac{1}{3}x^2-\dfrac{5}{2}-\dfrac{1}{6}\right)-2x^5-x^3\)
\(=6x^3\left(\dfrac{1}{3}x^2-\dfrac{16}{6}\right)-2x^5-x^3\)
\(=2x^5-16x^3-2x^5-x^3\)
\(=-17x^3\)
b) \(\left(x+3\right)\left(x^2+3x-2\right)\)
\(=x^3+3x^2-2x+3x^2+9x-6\)
\(=x^3+6x^2+7x-6\)
c) \(\left(4x^3-4x^2-5x+4\right):\left(2x+1\right)\)
\(=2x^2+4x^3-2x-4x^2-\dfrac{5}{2}-5x+\dfrac{2}{x}+4\)
\(=4x^3-2x^2-7x+\dfrac{2}{x}+\dfrac{3}{2}\)
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Làm tính chia :
a) \(\left(6x^3-7x^2-x+2\right):\left(2x+1\right)\)
b) \(\left(x^4-x^3+x^2+3x\right):\left(x^2-2x+3\right)\)
c) \(\left(x^2-y^2+6x+9\right):\left(x+y+3\right)\)
a) \(\left( {6{x^3} - 7{x^2} - x + 2} \right):\left( {2x + 1} \right)\)
b) $(x^4-x^3+x^2+3x):(x^2-2x+3)$
c) \(\left( {{x^2} + {y^2} + 6x + 9} \right):\left( {x + y + 3} \right)\)
\(=\left( {{x^2} + 6x + 9 - {y^2}} \right)\left( {x + y + 3} \right)\)
\(=\left[ {\left( {{x^2} + 2x.3 + {3^2}} \right) - {y^2}} \right]:\left( {x + y + 3} \right)\)
\(=\left[ {{{\left( {x + 3} \right)}^2} - {y^2}} \right]:\left( {x + y + 3} \right)\)
\(=\left( {x + 3 - y} \right)\left( {x + 3 + y} \right):\left( {x + y + 3} \right)\)
$= x + 3 - y$
$= x - y + 3$
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(6x3 - 7x2 - x + 2) : (2x + 1)
= (6x3 + 3x2 - 10x2 - 5x + 4x + 2) : (2x + 1)
= [(6x3 + 3x2) - (10x2 + 5x) + (4x + 2)] : (2x + 1)
= [3x2(2x + 1) - 5x(2x + 1) + 2(2x + 1)] : (2x + 1)
= (3x2 - 5x + 2)(2x + 1) : (2x + 1)
= 3x2 - 5x + 2
(x4 - x3 + x2 + 3x) : (x2 - 2x + 3)
= (x4 + x3 - 2x3 - 2x2 + 3x2 + 3x) : (x2 - 2x + 3)
= [(x4 + x3) - (2x3 + 2x2) + (3x2 + 3x)] : (x2 - 2x + 3)
= [x3(x + 1) - 2x2(x + 1) + 3x(x + 1)] : (x2 - 2x + 3)
= (x3 - 2x2 + 3x)(x + 1) : (x2 - 2x + 3)
= x(x2 - 2x + 3)(x + 1): (x2 - 2x + 3)
= x(x + 1)
= x2 + x
(x2 - y2 + 6x + 9) : (x + y + 3)
= [(x2 + 6x + 9) - y2] : (x + y + 3)
= [(x + 3)2 - y2] : (x + y + 3)
= (x + 3 + y)(x + 3 - y) : (x + y + 3)
= (x + y + 3)(x - y + 3) : (x + y + 3)
= x - y + 3
CHÚC BN HOK TỐT 
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c) (x2+y2+6x+9):(x+y+3)
= (x2+6x+9−y2)(x+y+3)
= [(x2+2x.3+32)−y2]:(x+y+3)
= [(x+3)2−y2]:(x+y+3)
= (x+3−y)(x+3+y):(x+y+3)
= x + 3 - y
= x - y + 3
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giải pt :
a,\(\left(6x-5\right)\sqrt{x+1}-\left(6x+2\right)\sqrt{x-1}+4\sqrt{x^2-1}=4x-3\)
b, \(\left(9x-2\right)\sqrt{3x-1}+\left(10-9x\right)\sqrt{3-3x}-4\sqrt{-9x^2+12x-3}=4\)
c, \(\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{-4x^2+16x-15}\)